Question

我使用以下代码来实现C中的基本产品 - 消费者问题。但在极少数情况下，我会得到错误的输出，如下所述。

Received值达到0，但当剩余生产者周期达到0时，接收器保持0值，输出出错。

请帮我理解我的代码有什么问题。感谢。

C代码：

#include <pthread.h>
#include <stdio.h>
#include <semaphore.h>
sem_t semaphore;
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
int queue[50];
int queueLength;
void *producer( void * param)
{
    for ( int i=0; i<50; i++ )
    {
        // Add item to queue
        pthread_mutex_lock( &mutex );
        queue[ queueLength++ ] = i;
        printf("Sent %d\n", i);
        pthread_mutex_unlock( &mutex );
        // Signal semaphore

        sem_post( &semaphore );
    }
}
void *consumer(void * param)
{
    for ( int i=0; i<50; i++ )
    {
        int item;
        // Wait if nothing in queue
        if (queueLength<1) { sem_wait(&semaphore); }
        pthread_mutex_lock( &mutex );
        item = queue[ -- queueLength ];
        pthread_mutex_unlock( &mutex );
        printf("Received %i\n", item);
    }
}
int main()
{
    pthread_t threads[2];
    sem_init( &semaphore, 0, 1 );
    pthread_create( &threads[0], 0, producer, 0 );
    pthread_create( &threads[1], 0, consumer, 0 );
    pthread_join( threads[0], 0 );
    pthread_join( threads[1], 0 );
    sem_destroy( &semaphore );
}

输出错误：

Sent 0
Sent 1
Sent 2
Sent 3
Sent 4
Sent 5
Sent 6
Sent 7
Sent 8
Sent 9
Sent 10
Sent 11
Sent 12
Sent 13
Sent 14
Sent 15
Sent 16
Sent 17
Sent 18
Sent 19
Sent 20
Sent 21
Sent 22
Sent 23
Sent 24
Sent 25
Sent 26
Sent 27
Sent 28
Received 28
Received 27
Received 26
Received 25
Received 24
Received 23
Received 22
Received 21
Received 20
Received 19
Received 18
Received 17
Received 16
Received 15
Received 14
Received 13
Received 12
Received 11
Received 10
Received 9
Received 8
Received 7
Received 6
Received 5
Received 4
Received 3
Received 2
Received 1
Received 0
Received 0
Received 0
Received 0
Received 0
Received 0
Received 0
Received 0
Received -8
Received 0
Received 0
Received 0
Received 0
Received 0
Received 0
Received 0
Received 0
Received 0
Received 0
Received 0
Received 0
Received 1
Sent 29
Sent 30
Sent 31
Sent 32
Sent 33
Sent 34
Sent 35
Sent 36
Sent 37
Sent 38
Sent 39
Sent 40
Sent 41
Sent 42
Sent 43
Sent 44
Sent 45
Sent 46
Sent 47
Sent 48
Sent 49

Answer 1

当您的消费者线程启动时，信号量已经发出信号，queueLength大于零。您的消费者线程将立即开始出列（因为queueLength不为零），直到它达到零。

一旦队列耗尽，在下一个消费者循环迭代中（使用queueLength == 0），它将输入if子句并尝试等待信号量（已经发出信号），并立即获得锁定。

所以信号量不能在if条件内，这意味着你可以使用类似的东西：

for ( int i=0; i<50; i++ )
{
    int item;

    // wait for signal
    sem_wait(&semaphore);

    pthread_mutex_lock( &mutex );
    item = queue[--queueLength];
    pthread_mutex_unlock( &mutex );
    printf("Received %i\n", item);
}

我的建议（如果这是你的实际问题）将是创建一个简单的循环缓冲区。如果正确实施（即只有一个生产者，单个消费者，如果生产者只移动head，而消费者只移动tail，并假设两个变量都是int或更小 - - 即可以原子方式读取和写入），你根本不需要mutex（只有信号量的信号量）。

您示例中名为queue的数组实际上是堆栈（LIFO），这在生产者/消费者模式中很常见，您通常希望先出列并处理最旧的项目。< / p>

使用C语言中的信号量和互斥锁进行生产者 - 消费者实现

1 个答案: